Elimination Method Calculator
Solve a system of linear equations by the elimination method — multiply equations by the right constants, add or subtract them to cancel out a variable, then back-substitute to find the rest. Works for 2 equations in 2 unknowns or 3 equations in 3 unknowns, with full step-by-step working and a visual for every case.
Background
The elimination method solves a system of linear equations without ever isolating a variable first (that's the substitution method's job). Instead, you scale the equations so that one variable's coefficients match, then add or subtract the equations so that variable cancels out entirely — leaving a smaller, simpler system to finish off the same way.
How to use this calculator
- Choose 2 Equations, 2 Unknowns for the classic case: two lines on the coordinate plane that either cross once, run parallel, or lie exactly on top of each other.
- Choose 3 Equations, 3 Unknowns to solve a larger system: elimination cascades down, clearing one variable at a time until only one is left, then builds the answer back up by substitution.
- Use Eliminate first to choose which variable the steps target first — the final answer is identical either way, since elimination works in any order.
- Click Calculate to see the classification (one solution, no solution, or infinitely many), a plain-language explanation, a supporting visual, and the full worked steps.
How the elimination method works
Pick a variable to eliminate, then multiply each equation by whatever constant makes that variable's coefficients match in size (opposite or equal signs both work, since you can add or subtract).
Add or subtract the two scaled equations so that variable cancels out completely, leaving one equation in the remaining variable(s).
Solve that smaller equation, then substitute the value back into any original equation to recover the eliminated variable.
For 3 equations, the same idea just cascades: eliminate one variable from two different pairs of equations to get two equations in two unknowns, then run the 2-variable elimination method on those.
If elimination ever produces "0 = 0," the equations were secretly dependent (multiples of each other) and there are infinitely many solutions. If it produces "0 = (a nonzero number)," the equations contradict each other and there's no solution at all.
The order you eliminate variables in never changes the final answer — it only changes which equation you end up solving first.
Formulas & Equations Used
2×2 system: a₁x + b₁y = c₁ and a₂x + b₂y = c₂
Eliminating y (multiply by b₂, b₁ and subtract): x = (c₁b₂ − c₂b₁) / (a₁b₂ − a₂b₁)
Eliminating x (multiply by a₂, a₁ and subtract): y = (a₁c₂ − a₂c₁) / (a₁b₂ − a₂b₁)
Classification: if the denominator D = a₁b₂ − a₂b₁ is 0, the system has no unique solution — check whether the numerators are also 0 (infinitely many solutions) or not (no solution).
3×3 system: eliminate one variable using Eq.1 & Eq.2, then again using Eq.1 & Eq.3, to get two equations in the remaining two variables — then apply the 2×2 method above, and back-substitute for the variable eliminated first.
Example Problems & Step-by-Step Solutions
Example 1 — Eliminating y first
2x + 3y = 13 and 4x − y = 5.
Step: Multiply the second equation by 3: 12x − 3y = 15. Add to the first: 14x = 28, so x = 2.
Result: Substituting back gives y = 3 — the point (2, 3).
Example 2 — Recognizing no solution
x + 2y = 4 and 2x + 4y = 10.
Step: Multiply the first equation by 2: 2x + 4y = 8. Subtract the second: 0 = −2.
Result: A false statement means no solution — these lines are parallel.
Example 3 — Recognizing infinitely many solutions
x + 2y = 4 and 2x + 4y = 8.
Step: Multiply the first equation by 2: 2x + 4y = 8. Subtract the second: 0 = 0.
Result: Always true, so there are infinitely many solutions — both equations describe the same line.
Example 4 — A 3-variable system
x+y+z=6, x−y+z=2, and 2x+y−z=1.
Step: Adding the first two eliminates y: 2x+2z=8, so x+z=4. Subtracting them gives y=2. Substituting into the third equation gives x=1, z=3.
Result: The solution is x=1, y=2, z=3 — check: 1+2+3=6 ✓.
Example 5 — A classic coin problem
30 coins (nickels, dimes, quarters) worth \$4.00 total, with twice as many dimes as nickels.
Step: n+d+q=30, d=2n, and 5n+10d+25q=400 (in cents). Substituting d=2n and eliminating q leaves a single equation in n.
Result: n=7, d=14, q=9 — check: 7+14+9=30 coins and 35¢+140¢+225¢=\$4.00 ✓.
Frequently Asked Questions
How is elimination different from substitution?
Substitution isolates one variable and plugs its expression into the other equation. Elimination never isolates anything — it scales and combines whole equations so a variable's terms cancel out on their own. Both always reach the same answer.
What if neither variable eliminates cleanly with small multipliers?
That's fine — the multipliers don't need to be small or even whole numbers. Cross-multiplying by the other equation's coefficient always works (multiply Eq.1 by Eq.2's coefficient and vice versa), even if the resulting numbers look messier.
Does it matter which variable I eliminate first?
No. You'll do slightly different arithmetic along the way, but you'll always land on the same final solution — or the same conclusion of no solution or infinitely many.
Why does "0 = 0" mean infinitely many solutions?
It means the second equation carried no new information beyond the first — it was really just a scaled copy. With only one genuine equation left for two (or more) unknowns, any point that satisfies it works, which is an entire line's worth of solutions.
Can a 3-variable system have exactly two solutions, or some other finite number greater than one?
No — a system of linear equations always has exactly one solution, no solution, or infinitely many. Two straight (flat) equations can never cross at exactly two points without being identical everywhere in between, so a finite count like "2" or "5" is never possible.
Why do 3-variable systems use elimination twice before solving anything?
Each elimination round removes one variable and one equation. Starting with 3 equations and 3 unknowns, one round of elimination (using two different pairs of equations) leaves 2 equations in 2 unknowns — exactly the case this calculator's other mode already solves.
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